Single command to replace breakpoint and open vscode debugger

[drisspg]

Summary

Often it can be pretty hard / tedious to replicate all the settings and env state when trying to debug through the integrated vscode config system. I have found it very convenient to just swap in the right breakpoint() on the line I want to debug.

Is it possible to have some equivalent that essentially is debugpy.breakpoint_and_attach() that encapsulates the full flow and can be used to drop into the debugger when hit? I tried to jerry-rig things up and didnt have much success, maybe this is already possible??

I use the integrated terminal

[rchiodo]

Transferred to discussions as not really an issue.

[rchiodo]

My first instinct here is can you just 'attach' to the process? That's essentially what you're asking for, you just want it to attach at the right spot. You stick a breakpoint in the code using the IDE ahead of time and after the attach it should stop on that breakpoint.

If you wanted to do this in code however, you'd put this code somewhere that you wanted it to 'wait' for the debugger:

import debugpy
debugpy.listen(("localhost", 5691))  # listen for incoming DAP client connections
debugpy.wait_for_client()  # wait for a client to connect

And then in a launch.json you'd have:

    {
      "name": "Python Debugger: Attach to listening debugpy",
      "type": "debugpy",
      "request": "attach",
      "connect": {
        "host": "localhost",
        "port": 5691
      },
    },

[drisspg]

Yeah this what I am currently doing, that being said it would be pretty magical if I didnt have to interact w/ the vscode debug UI at all until the debugger has launched

[rchiodo]

@karthiknadig did you have something that would be listening in VS code for a debug attach? That's what this would require. VS code doesn't currently have a 'listen' mode itself so an extension would have to be written to do so.